The airplane's speed relative to the ground is
√ (100² + 25²)
= √ (10,000 + 625)
= √ 10,625
= 103.08 km/hr .
The angle of its velocity north of west is
the angle whose tangent is (25/100)
arctan(25/100) = 14° north of west .
(bearing = 284°)
Answer:
Approximately 
if that athlete jumped up at 
. (Assuming that 
.)
Explanation:
The momentum 
of an object is the product of its mass 
and its velocity 
. That is: 
.
Before the jump, the speed of the athlete and the earth would be zero (relative to each other.) That is: 
and 
. Therefore:
and 
.
Assume that there is no force from outside of the earth (and the athlete) acting on the two. Momentum should be conserved at the instant that the athlete jumped up from the earth.
Before the jump, the sum of the momentum of the athlete and the earth was zero. Because momentum is conserved, the sum of the momentum of the two objects after the jump should also be zero. That is:
.
Therefore:
.
.
Rewrite this equation to find an expression for 
, the speed of the earth after the jump:
.
The mass of the athlete needs to be calculated from the weight of this athlete. Assume that the gravitational field strength is 
.
.
Calculate using 
and 
values from the question:
.
The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately .
Answer:
A) and B) are correct.
Explanation:
If the object is at rest, it means that no net force is exerted on it.
As the object experiences a downward gravitational force from Earth, in order to be at rest, it must experience an upward force with the same magnitude as the gravitational force on the object.
This force is supplied by the normal force, which can adopt any value in order to meet the condition imposed by Newton´s 2nd Law, and is always perpendicular to the surface on which the object is placed (in this case, the ground).
At a molecular level, this normal force is supplied by the bonded molecules of the ground that behave like small springs being compressed by the molecules of the object, exerting an upward restoring force upward on them.
So, the statements A) and B) are true.
Answer:
A: T = 120 N
B: T = 88.42 N
C: T = 70 N
Explanation:
Part A:
Since, the lighter bucket is supported by my had. So, the only unbalanced force in the system is the weight of heavier bucket. Hence, the tension in rope will be equal to the weight of heavier bucket.
<u>T = 120 N</u>
<u></u>
Part B:
This is the case where, two masses hang vertically on both sides of the pulley. To find the tension in such case we have the formula:
T = (2 m₁m₂g)/(m₁+m₂)
where,
m₁ = mass of heavier object = W₁/g = (120 N)/(9.8 m/s²) = 12.24 kg
m₁ = mass of lighter object = W₂/g = (70 N)/(9.8 m/s²) = 7.14 kg
g = 9.8 m/s²
Therefore,
T = [(2)(12.24 kg)(7.14 kg)(9.8 m/s²)]/(12.24kg + 7.14 kg)
T = 1713.6 N.kg/19.38 kg
<u>T = 88.42 N</u>
<u></u>
Part C:
Since, the heavier bucket is on ground. So, its weight is balanced by the normal reaction of the ground. The only unbalanced force in the system is the weight of lighter bucket. Hence, the tension in rope will be equal to the weight of lighter bucket.
<u>T = 70 N</u>
Answer:
Slip Rings
Explanation:
The wound rotor motor has a three-phase winding with each one connected to seperate slip rings. These slip rings contain brushes which form a secondary circuit where resistance can be inserted and this will allow for the rotor current to run more in phase with the stator current which will result in increased torque that is created
